Some Useful Trigonometric Limits

Before I discuss differentiating trigonometric functions, we should probably review a few key useful limits.

0. Useful Statements. We will use some statements that may or may not be obvious.

Squeeze Theorem. Suppose we have f(x)\leq g(x)\leq h(x) for all x\in (a,c). If b\in(a,c) and
\displaystyle L = \lim_{x\to b}f(x)=\lim_{x\to b}h(x)
\displaystyle \lim_{x\to b}g(x) = L.

1. Sinc Limit. We will consider a function (called the “sinc” function) \mathrm{sinc}(x)=x^{-1}\sin(x). What happens as x\to0? Lets think about it.

We will consider 0<x<\pi/2, then we know
(1)\displaystyle \sin(x)\leq x\leq\tan(x).
Divide through by \sin(x)
(2)\displaystyle 1\leq \frac{x}{\sin(x)}\leq\frac{\tan(x)}{\sin(x)}=\frac{1}{\cos(x)}.
Well, we know that
(3)\displaystyle \lim_{x\to0}\frac{1}{\cos(x)}=1.
Thus we have our limit bounded
(4)\displaystyle \lim_{x\to 0}1\leq\lim_{x\to0}\frac{x}{\sin(x)}\leq\lim_{x\to0}\frac{1}{\cos(x)}=1.
We rewrite this to stress the boundedness:
(4′)\displaystyle 1\leq\lim_{x\to0}\frac{x}{\sin(x)}\leq1.
This implies
(5)\displaystyle \lim_{x\to0}\frac{x}{\sin(x)}=1.
More importantly, this implies (taking the recipricol):
(6)\displaystyle \lim_{x\to0}\frac{\sin(x)}{x}=1.
This is one important limit.

2. Cosine Version. This is all very nice, but lets consider the limit
(7)\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=\textbf{??}
What to do? Well, we can do the following trick
(8)\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=\lim_{x\to0}\frac{1-\cos(x)}{x}\frac{1+\cos(x)}{1+\cos(x)}
The numerator (top of the fraction) becomes
\displaystyle (1-\cos(x))(1+\cos(x)) = 1 - \cos^{2}(x)=\sin^{2}(x).
We plug this in, rewriting our limit as
(9)\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=\lim_{x\to0}\frac{\sin^{2}(x)}{x(1+\cos(x))}.
What to do now?

We use the squeeze theorem, noting that the fraction has bounds
(10)\displaystyle   \frac{\sin^{2}(x)}{2x}\leq\frac{\sin^{2}(x)}{x(1+\cos(x))}\leq\frac{\sin^{2}(x)}{x}.
Using limit properties, we find that
(11)\displaystyle   \lim_{x\to0}\frac{\sin^{2}(x)}{2x}=\frac{1}{2}\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\sin(x)=(1)\cdot(0)=0.
Thus our problem has bounds
(12)\displaystyle 0\leq\lim_{x\to0}\frac{\sin^{2}(x)}{x(1+\cos(x))}\leq0.
The squeeze theorem produces the solution
(13)\displaystyle \lim_{x\to0}\frac{\sin^{2}(x)}{x(1+\cos(x))}=0.
Rewriting the function in familiar notation, we have
(13′)\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=0.

3. Punchline. So, we have just proven two important limits:
(14a)\displaystyle \lim_{x\to0}\frac{\sin(x)}{x}=1.
(14b)\displaystyle \lim_{x\to0}\frac{1-\cos(x)}{x}=0.

Exercise 1. Prove or find a counter-example: for any fixed constant c we have \displaystyle\lim_{x\to\infty} x\sin(c/x) = c.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
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3 Responses to Some Useful Trigonometric Limits

  1. Pingback: Differentiating Trigonometric Functions | My Math Blog

  2. Pingback: [Index] Calculus of a Single Variable | My Math Blog

  3. Pingback: [Index] Calculus in a Single Variable | My Math Blog

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